Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

<image>

In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int ara[103][103];
pair<int, int> coOrdinate;
int n, m;
bool RowCheck(int row) {
  for (int i = 1; i <= m; i++) {
    if (ara[row][i] == 0) return false;
  }
  return true;
}
bool ColCheck(int col) {
  for (int i = 1; i <= n; i++) {
    if (ara[i][col] == 0) return false;
  }
  return true;
}
int FindMax() {
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      if (ara[i][j] >= mx) {
        mx = ara[i][j];
        coOrdinate = make_pair(i, j);
      }
    }
  }
  return mx;
}
void GoRow(int row) {
  for (int i = 1; i <= m; i++) ara[row][i]--;
}
void GoCol(int col) {
  for (int i = 1; i <= n; i++) ara[i][col]--;
}
int main() {
  scanf("%d %d", &n, &m);
  int mx = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      scanf("%d", &ara[i][j]);
    }
  }
  mx = FindMax();
  int k = 0;
  vector<pair<string, int> > v;
  while (1) {
    if (mx != 0) {
      bool R = RowCheck(coOrdinate.first);
      bool C = ColCheck(coOrdinate.second);
      if (!R && !C)
        break;
      else if (R && !C) {
        GoRow(coOrdinate.first);
        v.push_back(make_pair("row", coOrdinate.first));
        k++;
      } else if (!R && C) {
        GoCol(coOrdinate.second);
        v.push_back(make_pair("col", coOrdinate.second));
        k++;
      } else {
        if (n < m) {
          GoRow(coOrdinate.first);
          v.push_back(make_pair("row", coOrdinate.first));
          k++;
        } else {
          GoCol(coOrdinate.second);
          v.push_back(make_pair("col", coOrdinate.second));
          k++;
        }
      }
    } else
      break;
    mx = FindMax();
  }
  if (mx != 0)
    cout << -1 << endl;
  else {
    cout << k << endl;
    int len = v.size();
    for (int i = 0; i < len; i++) {
      cout << v[i].first << " " << v[i].second << endl;
    }
    cout << endl;
  }
  return 0;
}